6

サブセレクトといくつかの集計関数を使用して、NULL および非 NULL 列を数えています。

CREATE TEMPORARY TABLE citizens(name text, country text,profession text,postalcode text);

INSERT INTO citizens VALUES
 ('Fred', 'USA', 'Professor', NULL),
 ('Amy', 'USA', 'Professor', NULL),
 ('Ted', 'USA', 'Professor', 90210),
 ('Barb', 'USA', 'Lawyer', 10248),
 ('Wally', 'USA', 'Lawyer', NULL),
 ('Fred', 'Canada', 'Professor', 'S0H'),
 ('Charles', 'Canada', 'Professor', 'S4L'),
 ('Nancy', 'Canada', 'Lawyer', NULL),
 ('Linda', 'Canada', 'Professor', NULL),
 ('Steph', 'France', 'Lawyer', 75008 ),
 ('Arnold', 'France', 'Lawyer', 75008 ),
 ('Penny', 'France', 'Lawyer', 75008 ),
 ('Harry', 'France', 'Lawyer', NULL);

SELECT country,
    profession,
    MAX(have_postalcode::int*num) AS num_have,
    MAX((1-have_postalcode::int)*num) AS num_not_have
FROM
    (
    SELECT country, profession,
        COUNT(*) AS num,
        (postalcode IS NOT NULL) AS have_postalcode
    FROM citizens
    GROUP BY country, profession, have_postalcode
    ) AS d
GROUP BY country, profession

結果とともに

USA     Professor   1   2
Canada  Lawyer      0   1
USA     Lawyer      1   1
France  Lawyer      3   1
Canada  Professor   2   1

しかし、より洗練された方法があるべきだと思われます (たとえばMAX、1 つの重要な値を取得するためだけに使用されるのは苦痛です)。誰かクールなアイデアを持っていますか？

4

3 に答える 3

8 
SELECT country, profession,
        COUNT(postalcode) AS num_have
      , (COUNT(*) - COUNT(postalcode)) AS num_not_have
FROM citizens
GROUP BY country, profession;

http://sqlfiddle.com/#!1/17a9d/15

于 2012-10-11T17:05:04.817 に答える
4 
SELECT  country
,       profession
,       sum(case when postalcode is not null then 1 end) as num_have
,       sum(case when postalcode is null then 1 end) as num_not_have
FROM    citizens
GROUP BY 
        country
,       profession
于 2012-10-11T16:58:10.867 に答える
4 
SELECT Country, Profession, 
    count(Country) as num_have, count(*) - count(PostalCode) as num_not_have
FROM citizens
GROUP BY Country, Profession
于 2012-10-11T16:59:03.860 に答える